How Do You Solve Second Derivatives Using Implicit Differentiation?

[AdiV]

Hi, I need some help with these question, and would appreciate the help.

Homework Statement

Part 1
Use implicit differentiation to find y'' if 2xy = y^2 Simplify and Leaev in terms of x and y.

Part 2
Use implicit differentiation to find y '' if xy + y^3 = 1
Simplify your answer and leave it in terms of x and y

Homework Equations

The Attempt at a Solution

For part 1;

I had

2x dy/dx + 2y = 2y dy/dx
Took out 2's
solved for dy/dx

dy dx = y / [y-x]

Part 2
I tried the same, took d/dx to get

dy/dx + 3y^2 dy/dx = 0 But here I get stuck because of the zero.

 

[rocomath]

It asked for y'' not y', let me work it but am I correct on that part?

 

[AdiV]

Ohh, so go through the process again? Ok, but can you help me along?

 

[rocomath]

Ohh, so go through the process again? Ok, but can you help me along?

Sure, so now we're at

$$y'=\frac{x}{y-x}$$

Take the derivative again, quotient rule:

$$y'=\frac{d}{dx}(\frac{x}{y-x})$$

so...

$$y''=\frac{(y-x)\frac{d}{dx}(x)-x\frac{d}{dx}(y-x)}{(y-x)^2}$$

 

[rocomath]

For part 2: You forgot or did the the product rule incorrectly;

$$xy+y^3 =1$$

$$xy'+y+3y^2 y'=0$$

The zero means nothing. It's the same thing as saying x-2=0, well then x=2.

Solving for y'

$$y'=\frac{-y}{x+3y^2}$$

Now take the 2nd derivative, quotient rule:

 

[HallsofIvy]

You don't need to solve for y' itself. If $2xy= y^2$, then $2y+ 2xy'= 2yy'$. Now differentiate both sides of that with respect to x: $2y'+ 2y'+ 2xy"= 2y'^2+ yy"$ or $4y'- 2y'^2= (y- 2x)y"$ so $y"= (4y'-2y'^2)/(y- 2x)$. Personally, I would consider that a perfectly good answer but since your problem specifically says "leave in terms of x and y", NOW use x/(y- x).

 

[AdiV]

Wow, thanks for clearing it up for me, I greatly appreciate the help =]